In - ABC prove that cos A - B-2 - a - b-csinc-2

cos(A B2) = (a + bc) #160; sinc2 L.H.S ⇒cos(A B2) sin(A + B2)sin(A + B2) = 1/2 (sin A + sin B)sin(A C2) ⇒12sin A + sin BcosC2sin C/2sin C/ ...

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Byju's Answer

Standard VII

Mathematics

Angle Sum Property

In Δ ABC pr...

Question

In $Δ A B C$ prove that
$cos (\frac{A - B}{2}) = (\frac{a + b}{c}) sin \frac{c}{2}$

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Solution

$cos (\frac{A - B}{2}) = (\frac{a + b}{c}) sin \frac{c}{2}$
L.H.S
$\Rightarrow \frac{cos (\frac{A - B}{2}) sin (\frac{A + B}{2})}{sin (\frac{A + B}{2})} = \frac{\frac{1}{2} (sin A + sin B)}{sin (\frac{A - C}{2})}$
$\Rightarrow \frac{1}{2} \frac{sin A + sin B}{cos \frac{C}{2}} \frac{sin C / 2}{sin C / 2}$
$\Rightarrow \frac{sin A + sin B}{sin C} sin \frac{C}{2}$
Using sin e law
$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c} = \frac{1}{2 R}$
$sin A = \frac{a}{2 R}, sin B = \frac{b}{2 R}, sin C = \frac{c}{2 R}$
$\Rightarrow \frac{a + b}{C} \frac{sin C}{2} =$ R.H.S
Hence proved

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In ΔABC prove thatcos(A−B2)=(a+bc)sinc2

cos(A−B2)=(a+bc)sinc2L.H.S⇒cos(A−B2)sin(A+B2)sin(A+B2)=12(sinA+sinB)sin(A−C2)⇒12sinA+sinBcosC2sinC/2sinC/2⇒sinA+sinBsinCsinC2Using sin e lawsinAa=sinBb=sinCc=12RsinA=a2R,sinB=b2R,sinC=c2R⇒a+bCsinC2= R.H.SHence proved

In $Δ A B C$ prove that
$cos (\frac{A - B}{2}) = (\frac{a + b}{c}) sin \frac{c}{2}$